If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? lead metal can be extracted from a mineral called galena, which contains 86.6% lead by mass. a particular ore contains 68.5% galena by mass.

The volume of sphere can be calculated using the following formula:

Here, r is radius of the sphere which is 3.5 cm. Putting the value,

This is equal to the volume of lead, density of lead is thus, mass of lead can be calculated as follows:

m=d×V

Putting the values,

Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.

Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:

m=0.686×0.866=0.5940 g

Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

1 g of lead obtain from grams of ore.

Thus, 2041.2 g of lead obtain from:

Therefore, mass of ore required to make lead sphere is 3.715 kg.

imma come back to you hold on

it's filtration, but this is an assumption so my apologies if i got this